What is the integral of tan^3(x/2) ?

The integral of tan^3(x/2) is -2ln|sec(x/2)|+sec^2(x/2)+C

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integral of tan^3(x/2)

\int tan^{3} (\frac{x}{2}) d x

- 2 ln sec (\frac{x}{2}) + sec^{2} (\frac{x}{2}) + C

Solution steps

\int tan^{3} (\frac{x}{2}) d x

Apply u-substitution

: \int tan^{3} (u) \cdot 2 d u

= \int tan^{3} (u) \cdot 2 d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int tan^{3} (u) d u

Rewrite using trig identities

= 2 \cdot \int (- 1 + sec^{2} (u)) tan (u) d u

Apply u-substitution

: \int \frac{- 1 + v ^{2}}{v} d v

= 2 \cdot \int \frac{- 1 + v ^{2}}{v} d v

Expand

\frac{- 1 + v ^{2}}{v} : - \frac{1}{v} + v

= 2 \cdot \int - \frac{1}{v} + v d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 (- \int \frac{1}{v} d v + \int v d v)

\int \frac{1}{v} d v = ln ∣ v ∣

\int v d v = \frac{v ^{2}}{2}

= 2 (- ln ∣ v ∣ + \frac{v ^{2}}{2})

Substitute back

= 2 (- ln sec (\frac{x}{2}) + \frac{sec ^{2} ( \frac{x}{2} )}{2})

Simplify

2 (- ln sec (\frac{x}{2}) + \frac{sec ^{2} ( \frac{x}{2} )}{2}) : - 2 ln sec (\frac{x}{2}) + sec^{2} (\frac{x}{2})

= - 2 ln sec (\frac{x}{2}) + sec^{2} (\frac{x}{2})

Add a constant to the solution

= - 2 ln sec (\frac{x}{2}) + sec^{2} (\frac{x}{2}) + C

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What is the integral of tan^3(x/2) ?
The integral of tan^3(x/2) is -2ln|sec(x/2)|+sec^2(x/2)+C